1-D Perfectly Elastic Collison PLEASE CHECK THANKS

AI Thread Summary
The discussion centers on a 1-D perfectly elastic collision problem involving two masses, where mass m1 (0.3155 kg) moves at 17.6 m/s and collides with a stationary mass m2 (0.685 kg). After the collision, m2 moves at 11.1 m/s, while m1's velocity is calculated to be -6.5 m/s. The total momentum before and after the collision is confirmed to be equal at 5.5 kg m/s, validating the calculations. Participants express a desire for confirmation of the solution's accuracy, emphasizing the importance of checking both momentum and kinetic energy conservation. The consensus is that the calculations are correct, as there is only one solution to the problem.
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1-D Perfectly Elastic Collison PLEASE CHECK! THANKS

Homework Statement


Mass m1 is moving to the right at a velocity of 17.6m/s. Suddenly it strikes a stationary ball. The stationary ball has a mass of .685kg. The collision is perfectly elastic and 1 dimensional. The collision forces m2 to move to the right at 11.1m/s (v2prime).

Total momentum = 5.5 kg m/s ??
m1 = .3155kg ??
v1prime= -6.5m/s ??

Homework Equations



momentum before = momentum after
m1v1 = m1v1prime + m2v2prime

v2prime = (2m1)/(m1+m2) (v1) + (m2-m1)/(m1+m2) (v2)

v1prime = (m1-m2)/(m1+m2) (v1) + (2m2)/(m1+m2) (v2)

The Attempt at a Solution

11.1 = (2*m1)/(m1+.685) ( 17.6) + 0 (cancels cause m2 is stationary
11.1(m1+.685) = 35.2m
11.1m1 + 7.6035 = 35.2m
24.1m = 7.6035
m1= .3155 kg

v1prime = (.316-.685)/(.316+.685)(17.6) + 0 (v2 = 0)
v1prime = -6.50 m/s Left.

momentum before = momentum after
(17.6)(.316) = (.316)(-6.5) + (.685)(11.1)
5.55 kgm/s


Thanks for taking your time to help ! I just want to make sure this is right.
 
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yeah, it's right. to check just see if the momentum is the same before & after and check if the kinetic energy is the same before & after. they are, so you did it right.
 


sorry to ask this, but are you 100% sure? Maybe I can get some more people to assure this answer is correct, as there are multiple other ways of doing the same problem.
 


proxy98 said:
sorry to ask this, but are you 100% sure? Maybe I can get some more people to assure this answer is correct, as there are multiple other ways of doing the same problem.

Yes. there is only one solution to the problem, so if you plug it into find the correct results, it must be right.
 
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